# Spectrum of a field

Let's $F$ be a field. What is $\operatorname{Spec}(F)$? I know that $\operatorname{Spec}(R)$ for ring $R$ is the set of prime ideals of $R$. But field doesn't have any non-trivial ideals.

Thanks a lot!

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It only has the $0$ ideal. – azarel Jul 27 '12 at 19:28
Just as Wikipedia says. – anon Jul 27 '12 at 19:29

As you say $\mathrm{Spec}(R)$ is defined to be the set of all prime ideals of $R$. If $R$ is a field, the only proper ideal is $0$ hence you get $\mathrm{Spec}(F) = \{0\}$.

It gets more interesting if your space is a ring that is not a field, like for example $R = \mathbb Z$. Then you can endow it with the following topology: each closed set in the space corresponds to an ideal $J$ of $R$, defined as $C(J) = \{ p \mid p \text{ a prime ideal of } R \text{ such that } J \subset p\}$.

Now what does $\mathrm{Spec}(\mathbb Z)$ endowed with this topology look like? Well, first of all, the points in our space correspond to prime ideals and since $\mathbb Z$ is a principal ideal domain, each point looks like $p\mathbb Z$. Note that the zero ideal $\{0\}$ is prime if and only if the ring is an integral domain, so in this case, zero is also a point in our space.

Next we want to know what closed sets look like. For this, let's stick a prime ideal into $C(\cdot)$ and see what comes out: $C(p\mathbb Z) = \{ p \mathbb Z \}$ which means, every singleton set in our space is closed.

Now what are the closed sets corresponding to non-prime ideals? Well, $n$ has only a finite number of prime divisors and each point in $C(n\mathbb Z)$ corresponds to a divisor of $n$: $C(n\mathbb Z) = \{ p \mathbb Z \mid p \mathbb Z \text{ a prime ideal containing } n \mathbb Z \}$.

Now we know that all closed sets are finite.

Edit (I apologise for the blunder kindly pointed out by Rene and t.b.)

You need to be careful about what open sets, i.e. complements of closed sets look like. You can easily trick yourself into believing that since a set is closed if and only if it's finite, $\mathrm{Spec}(\mathbb Z)$ has the cofinite topology. But this is false. To see this note that if we indeed had the cofinite topology, $\mathrm{Spec}(\mathbb Z) \setminus \{\langle 0 \rangle \}$ would be open. But for this to be true, $\langle 0 \rangle$ would have to be closed which means that we would have to have an ideal $I$ in $\mathbb Z$ such that the only prime ideal it is contained in is $\langle 0 \rangle$. But this implies that $I = \langle 0 \rangle$ which implies that $I$ is contained in every prime ideal. Hence there is no ideal $I$ such that $C(I) = \{ \langle 0 \rangle \}$.

As pointed out in Rene's answer, it boils down to all open sets contain zero since the complement of a closed set, $C(n\mathbb Z)^c$, is all prime ideals contained in $n \mathbb Z$ which always includes zero since we're in an integral domain so that the zero ideal is prime.

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There's another answer now, pointing out that you missed the generic point $0$ on $\operatorname{Spec}\mathbb{Z}$ which is contained in every open set. – t.b. Jul 28 '12 at 13:16
@t.b. Thanks for the ping. – Rudy the Reindeer Jul 28 '12 at 14:17

$Spec(Z)$ does not have the cofinite topology. The non-empty open sets are precisely the cofinite sets that contain the zero ideal.

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Thank you Rene, +1! – Rudy the Reindeer Jul 28 '12 at 14:18

I know that the spectrum of a field in the Zariski/commutative-algebra/algebraic-geometry meaning is simply $\{ (0) \}$ but it is too trivial, so I suspect there is another meaning for this term applied to fields. In a paper by Yuval Flicker et al. there is a related term: "real spectrum of a field" $\Omega_k$ which is the topological space of all orderings $\xi$ of $k$ ($k$ is a field). Its topology is generated by the subsets $$\{ \xi \in \Omega_k \mid a > 0 \mbox{ at } \xi \}, \quad a \in k \ .$$ So I ask: is there a generalization of this which is called spectrum of a field?

See: Yuval Z. Flicker, Claus Scheiderer, R. Sujatha, Grothendieck's Theorem on Non-Abelian $H^2$ and Local-Global Principles, Journal of the American Mathematical Society, 1998*.

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